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4 years ago

Determine whether AB←→ and CD←→− are parallel, perpendicular, or neither. A(−1, −4), B(2, 11), C(1, 1), D(4, 10)

1 answer:

4 0

Slope of AB
m = (y2 - y1)/(x2 - x1)
m = (11 - (-4))/(2 - (-1))
m = (11 + 4)/(2 + 1)
m = 15/3
m = 5
The slope of line AB is 5

Slope of CD
m = (y2 - y1)/(x2 - x1)
m = (10 - 1)/(4 - 1)
m = 9/3
m = 3
The slope of CD is 3

Since the two slopes are not equal, this means the lines are not parallel.
Since (slope of AB)*(slope of CD) = 5*3 = 15 is not equal to -1, this means the lines are not perpendicular

Answer: Neither (not parallel; not perpendicular)

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3 years ago

Read 2 more answers

evaluate the line integral ∫cf⋅dr, where f(x,y,z)=5xi−yj+zk and c is given by the vector function r(t)=⟨sint,cost,t⟩, 0≤t≤3π/2.

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We have

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} \vec f(\vec r(t)) \cdot \dfrac{d\vec r}{dt} \, dt$

and

$\vec f(\vec r(t)) = 5\sin(t) \, \vec\imath - \cos(t) \, \vec\jmath + t \, \vec k$

$\vec r(t) = \sin(t)\,\vec\imath + \cos(t)\,\vec\jmath + t\,\vec k \implies \dfrac{d\vec r}{dt} = \cos(t) \, \vec\imath - \sin(t) \, \vec\jmath + \vec k$

so the line integral is equilvalent to

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (5\sin(t) \cos(t) + \sin(t)\cos(t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (6\sin(t) \cos(t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (3\sin(2t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \left(-\frac32 \cos(2t) + \frac12 t^2\right) \bigg_0^{\frac{3\pi}2}$

$\displaystyle \int_C \vec f \cdot d\vec r = \left(\frac32 + \frac{9\pi^2}8\right) - \left(-\frac32\right) = \boxed{3 + \frac{9\pi^2}8}$

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